2g of benzoic acid $(C_{6} H_{5} COOH)$ dissolved in 25g of benzene shows a depression in freezing point equal to 1.62K. Molal depression constant for benzene is 4.9K Kg mol^-1. The percentage association of acid if it forms dimer in solution is

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9.92%

90.08%

99.2%

0.8%

answer is B.

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Detailed Solution

$\begin{aligned} {ΔT}_{f} = K_{f} \times \frac{W_{B}}{M_{B}} \times \frac{1000}{W_{A}} \\ M_{B} = \frac{4.9 \times 2 \times 1000}{1.62 \times 25} = 241.98 g / mol \end{aligned}$ $Total number of moles of particles \Rightarrow 1 - x + x / 2 = 1 - x / 2 = i \Rightarrow i = \Rightarrow 1 - x / 2 = 122 / 241.98 \Rightarrow x / 2 = 1 - 122 / 241.98 \Rightarrow x = 0.992 = 99.2 % ∴ Degree of association of benzene = 99.2 %$